The curve which extremizes I=int0^pi/4[(y")^2-y^2+x^2]dx under y(0)=0, y'(0)=1, y(pi/4)=y'(pi/4)=frac1sqrt2 is:

Q60Section II (Mathematics)60 of 145

$y^{(4)}=y$ , and $\sin x$ meets all four end conditions

The curve which extremizes $I=\int_0^{\pi/4}\bigl[(y")^2-y^2+x^2\bigr]dx$ under $y(0)=0$ , $y'(0)=1$ , $y(\pi/4)=y'(\pi/4)=\frac1{\sqrt2}$ is:

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